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Yeah that's very good but then in order to find the value of n we would've to find the value of c/a which are the coefficients in f(x). 

Anyway, thanks a lot catmg! :D Let me see what can be done further

(14x+22)/(2x+18)=3x+1       Multiply both sides of the equation by (2x+18)

14x +22 = (3x+1)(2x+18)     expand and simplify to

6x^2+42x-4 = 0     Now use Quadratic Formula       to find x values    

       a = 6    b = 42    c = -4    in the Quadratic Formula

\(x = {-42 \pm \sqrt{42^2-4(6)(-4)} \over 2(6)}\)

Woah, this is a hard problem, I can't figure it out, but maybe this is helpful. 

f(x) = a(x-1)^4 - 1, where a is at most 3 degrees. 

f(x) = b(x+1)^4 + 1, where b is at most 3 degrees. 

This would mean that f(1) = -1, f(-1) = 1. 

The sum of the coeficients is -1, and the even degree coefficients - odd degree coefficients = 1. 

By vieta's, ax^(n) + bx^(n-1) + cx^(n-2)..., the pairwise product of all the roots would be c/a. 

Good luck. :))

=^._.^=

https://web2.0calc.com/questions/help-geometry_119#r1

Find the coefficient of ...what ?

The coefficient is the constant in front of the variable ....the RED numbers below are examples of coefficients:

2x

3y^2

1/4 z^5

x = total pupils

1/4x +3     +   1/3(x - 1/4 x -3) -3      + 21    =   x

1/4x + 1/3x - 1/12x + 3-1-3+21  = x

1/2 x +20 = x

1/2x = 20

x = 40

Triangle side length rule says the sum of any two sides must be greater than the remaining third side

60 - 25 = 35 inches for the remaining two sides

25    6    29     any less ( like 5  30 )   would violate the rule   ( because 5 + 25 would not be greater than the '30' leg)

Melody,

From what I understand it says that f(x) is a polynomial of degree atmost 7 right? then it would have atmost 7 roots.

Let's then consider its roots are a,b,c,d,e,f,g then, sum of pairwise product of all roots means ab + bc + cd + de + ef + fg + ag + bg + cg +...+ eg + af + bf +.... so on = n you get the point right?

I think then we would've to find that sum to find the value of n. 

Another thing, its also given that f(x) leaves remainder -1 and 1 on division with (x-1)⁴ and (x+1)⁴ I don't know what to deduce from that...

Hi Guest,

It has asked for a three digit, odd number.

Not a number with 3 odd digits.   

That is how EP and I have interpreted it anyway.

Thanks EP and also guest.

EP has done what I suggested.

there are 2 numbers, (1 and 3) that could go in the units place

then there are 4 that can go in the tens and 4 that can go in the hundreds

2*4*4 = 32

I have not answered it but I have made a post.

It would be good if another mathematician explains it to us both.   

I actually do not understand what this means

"If the sum of pairwise product of all roots of  f(x)=0  is n"    Is it even meant to be real like that??

I thought it meant that if you find the product of every pair of roots and add them together then the answer is 0. 

But that would mean that n has no meaning in the problem. 

So I am really confused.

Amy, do you understand what the question is asking?  If so please explain it to me.   

The correct answer is this:

A / [XYZ] = 851 / 999  

[PAB] / [PBR] = 2 / 7

In the diagram below
\(AM = BM = CM\), and
\(\angle BMC+\angle A = 210^\circ\).
Find \(\angle B\) in degree

\(\begin{array}{|rcll|} \hline \angle BMC+\angle A &=& 210^\circ \\ \mathbf{\angle BMC} &=& \mathbf{210^\circ-\angle A} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \text{In }\triangle AMC \\ \hline (180^\circ-\angle BMC)+2*(\angle A) &=& 180^\circ \\ 2*(\angle A)&=& \angle BMC \quad | \quad \mathbf{\angle BMC=210^\circ-\angle A}\\ 2*(\angle A) &=& 210^\circ-\angle A \\ 3*(\angle A) &=& 210^\circ \\ \mathbf{\angle A} &=& \mathbf{70^\circ} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \text{In }\triangle BMC \\ \hline \angle BMC+2*(\angle B) &=& 180^\circ \quad | \quad \angle BMC = 2*(\angle A) \\ 2*(\angle A)+2*(\angle B) &=& 180^\circ \\ \angle A+\angle B &=& 90^\circ \\ \angle B &=& 90^\circ -\angle A \\ \angle B &=& 90^\circ -70^\circ \\ \mathbf{\angle B} &=& \mathbf{20^\circ} \\ \hline \end{array}\)

Question: Compute

\(\dfrac{5}{4 \cdot 9} + \dfrac{7}{9 \cdot 16} + \dfrac{9}{16 \cdot 25} + \dfrac{11}{25 \cdot 36}\)

\(\begin{array}{|rcll|} \hline && \mathbf{\dfrac{5}{4 \cdot 9} + \dfrac{7}{9 \cdot 16} + \dfrac{9}{16 \cdot 25} + \dfrac{11}{25 \cdot 36}} \\\\ &=& \dfrac{5}{9 \cdot 4} + \dfrac{7}{16 \cdot 9} + \dfrac{9}{25 \cdot 16} + \dfrac{11}{36 \cdot 25} \\\\ &=& \dfrac{9-4}{9 \cdot 4} + \dfrac{16-9}{16 \cdot 9} + \dfrac{25-16}{25 \cdot 16} + \dfrac{36-25}{36 \cdot 25} \\\\ &=&\small{ \dfrac{9}{9 \cdot 4} -\dfrac{4}{9 \cdot 4} + \dfrac{16}{16 \cdot 9} -\dfrac{9}{16 \cdot 9} + \dfrac{25}{25 \cdot 16} -\dfrac{16}{25 \cdot 16} + \dfrac{36}{36 \cdot 25} -\dfrac{25}{36 \cdot 25}} \\\\ &=& \dfrac{1}{4} -\dfrac{1}{9} + \dfrac{1}{9} -\dfrac{1}{16} + \dfrac{1}{16} -\dfrac{1}{25} + \dfrac{1}{25} -\dfrac{1}{36} \\\\ &=& \dfrac{1}{4} -\dfrac{1}{36} \\\\ &=& \dfrac{36-4}{4*36} \\\\ &=& \dfrac{32}{4*36} \\\\ &=& \dfrac{8}{36} \\\\ &=& \mathbf{\dfrac{2}{9}} \\ \hline \end{array}\)

\(\dfrac{5}{4 \cdot 9} + \dfrac{7}{9 \cdot 16} + \dfrac{9}{16 \cdot 25} + \dfrac{11}{25 \cdot 36} = \mathbf{\dfrac{2}{9}}\)

https://web2.0calc.com/questions/domain_100

log_3 (x - 1) = -1

means:

 (x-1) = 3^-1

x-1 = 1/3

x = 4/3 

DC = CE = DE, so CE = CB (since its  a square). 90 - 60 = 30. 180 - 30 = 150. 150 / 2 = 75.

The witdh of the interval is 16.

Using the discriminant, the largest possible value of k is 36.

What is A + B?

Hello Guest!

The function is indefinite for the x-values of the equation x ^ 2 - 6x + 4 = 0.

\(x ^ 2 - 6x + 4 = 0\\ x=3\pm\sqrt{9-4}\\ x=3\pm\sqrt{5}\)

\(A=3+\sqrt{5}\\ B=3-\sqrt{5}\\ A+B=6\)

  !

(kn^2)/(lm) = (6*2^2)/(2*1^2) = 12.